Posted by Subhotosh Khan on August 19, 2002 at 09:08:46:
In Reply to: Re: Have you learnt about trigonometric substitutions - yet? (n/t) posted by Soroban on August 15, 2002 at 23:21:06:
: : : Can anyone help me with both an answer AND an explanation.
: : : INTEGRATE: 1 / x((x^2 - 4)^1/2)
: : : and
: : : INTEGRATE: 1 / (x^1/2)(1 - 2(x^1/2))
: : : Thank you for taking the time.
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: Sorry, Subhotosh... your own fractions confused you.
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I think not- because the problem was (unless I am counting the parentheses wrong):
INT[1 / x((x^2 - 4)^1/2)]
=INT[1 / x * ((x^2 - 4)^1/2)]
=INT[((x^2 - 4)^1/2) / x]
If the problem was:
INT[1 / {x ((x^2 - 4)^1/2)}]
Then I would agree with you...
: Once again, a thousand pardons!
: The denominator is: x^(1/2) - 2x, which requires some work.
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No. The problem was
1 / (x^1/2) (1 - 2(x^1/2))
And NOT
1 / {(x^1/2) (1 - 2(x^1/2))}
so:
1 / (x^1/2) (1 - 2(x^1/2))
=(1 - 2(x^1/2)) / (x^1/2)
The problem COULD have been MEANT to be the way you showed - but it was surely not written that way...
I respect your knowledge and I know that my fingers get ahead of my head many times. I depend on others to show my mistakes - that way I can avoid bigger one.
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